Why is $e^{it}$ a submersion, and what is the relationship between the derivative $\dot h(t)$ and the differential $h_{*,t}$?

Question

My book is An Introduction to Manifolds by Loring W. Tu.

In the paragraph below, why exactly is $h: \mathbb R \to S^1, h(t)=(\cos t, \sin t) \cong e^{it}$ a submersion, and what is the relationship between the derivative $\dot h(t) = (-\sin t, \cos t) \cong ie^{it}$ and the differential $h_{*,t}$?

My argument is that the image of the differential at any $t$, $h_{*,t}: T_t(\mathbb R) \to T_{e^{it}}(S^1)$, is a vector subspace of $S^1$, which has dimension 1, so the image has dimension either $0$ or $1$, so the differential is either trivial or surjective. I think I can show each $h_{*,t}$ is not trivial directly by computing the image:

$$im(h_{*,t}) = h_{*,t}(T_t(\mathbb R)) = \{h_{*,t}(a \frac{d}{dt})\}_{a \in \mathbb R} = \{ah_{*,t}(\frac{d}{dt})\}_{a \in \mathbb R}$$

$$\ne \{\text{the zero element of} \ T_{e^{it}}(S^1) \ \text{which I think corresponds to the point} \ 1+0i \in S^1 \}$$

I cannot seem to identify the precise relationship between the derivative $\dot h(t)$ and the differential $h_{*,t}$, which I think I would assume in proving the preceding inequality. For a function $f: M \to \mathbb R$, there is a relationship between $f$'s a submersion at $p$ and $f$'s partial derivatives, but now we have $f: \mathbb R \to M$.

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Update: My answer considers the inclusion $g = \iota \circ h$ of $h$ into $\mathbb R^2$. After all, we have in the first place that $h$ is smooth by Theorem 11.15 because $g$ is smooth.

1. For such inclusion $g = \iota \circ h: \mathbb R \to \mathbb R^2$, $g(t_0) = (\iota \circ h)(t_0) = \iota(h(t_0))=h(t_0), \iota: S^1 \to \mathbb R^2$, we have by chain rule that for each $t_0 \in \mathbb R$, $g_{*,t_0}: T_{t_0}\mathbb R \to T_{h(t_0)}\mathbb R^2$ is given by $g_{*,t_0} = \iota_{*,h(t_0)} \circ h_{*,t_0}$.

2. For each $t_0 \in \mathbb R$, $\iota_{*,h(t_0)}: T_{h(t_0)}S^1 \to T_{h(t_0)}\mathbb R^2$ is still inclusion so $\iota_{*,h(t_0)} \circ h_{*,t_0} = h_{*,t_0}$

3. Let $t_0 \in \mathbb R$. If $h_{*,t_0}$ is not the zero map, then $h_{*,t_0}$ is surjective because the image of $h_{*,t_0}: T_{t_0}\mathbb R \to T_{h(t_0)}S^1$ is a subspace of $T_{h(t_0)}S^1 \cong \mathbb R$ and thus is (isomorphic to) either $\mathbb R$ or $\mathbb R^0 = \{0\}$.

4. Now, I'll argue that $h_{*,t_0}$ is not the zero map if I can show that $g_{*,t_0}$ is not the zero map since $T_{h(t_0)}S^1$ and $T_{h(t_0)}\mathbb R^2$ share the same zero element since $T_{h(t_0)}S^1$ is a vector subspace of $T_{h(t_0)}\mathbb R^2$.

5. $g_{*,t_0}$ is not the zero map if the geometric derivative (or velocity vector) $g'(t_0) := g_{*,t_0}[\frac{d}{dt}|_{t_0}]$ is not the zero vector i.e. if $g_{*,t_0}$ maps the basis element of $T_{t_0}\mathbb R$ to a nonzero element of $T_{h(t_0)}\mathbb R^2$.

6. Finally, $g'(t_0)$, by Proposition 8.15, is represented by calculus derivative $\dot g(t) = (-\sin(t),\cos(t))$ through $g'(t_0) = -\sin(t)\frac{\partial}{\partial r^1} + \cos(t) \frac{\partial}{\partial r^2}$, where $r^1$ and $r^2$ are the standard coordinates on $\mathbb R^2$ such that $\{\frac{\partial}{\partial r^1}, \frac{\partial}{\partial r^2} \}$ forms a basis for $T_{h(t_0)}\mathbb R^2$.

7. By (6), $g'(t_0)$ is not the zero vector of $T_{h(t_0)}\mathbb R^2$.

8. Therefore, by (7),(5),(4) $g_{*,t_0}$ is the not the zero map.

9. Therefore, by (3) and (8), neither is $h_{*,t_0}$ i.e. $h'(t_0)$ is the not the zero vector of $T_{h(t_0)}S^1$.

Answer 1

In the 1-dimensional vector space $T_t(\mathbb R)$ the standard basis element, as you seem to have noticed, is $\frac{d}{dt}$.

At $(\cos(t),\sin(t))$, the 1-dimensional vector space $T_{(\cos(t),\sin(t))} S^1$ has standard basis element $(-\sin(t),\cos(t))$.

In both cases, by "standard" I mean that the basis element has length 1 with respect to the "usual" norm, and it points in the direction of the "standard" orientation.

The relationship between $\dot h(t)$ and the differential $h_{*,t}$ is $$h_{*,t}(a \frac{d}{dt}) = a h_*(\frac{d}{dt}) = a (-sin(t),\cos(t)) = (-a\sin(t),a\cos(t)) $$

Answer 2

With regards to the difference between $h'(t_0)$ and $h_{*,t_0}$, $$ h_{*,t_0}\frac{d}{dt}=\frac{d}{dt}(h)\bigg|_{t_0}=h'(t_0).$$ In particular, the differential map $h_{*,t_0}$ is a linear map $h_{*,t_0}: T_{t_0} \mathbb{R}\to T_{h(t_0)}S^1$. $h'(t_0)$ is the image of the tangent vector $$ \frac{d}{dt}\bigg|_t\in T_t\mathbb{R}.$$ As for this problem: because we have a linear map $h_{*,t_0}$ between one-dimensional vector spaces the image is all of the vector space so long as the transformation does not send $d/dt|_{t_0}\in T_{t_0}\mathbb{R}$ to $0$. For $t_0$, we have $$ h_{*,t_0}\frac{d}{dt}\bigg|_{t_0}=h'(t_0)=(-\sin t_0,\cos t_0).$$ A brief argument will show that $\sin t$ and $\cos t$ are never simultaneously zero. So, $h'(t_0)$ is never the zero vector, hence $h_*$ is always surjective, so $h$ is a submersion.